Public key exponent coprime with totient proof

Question

I read this question a while back (In RSA, why is it important to choose e so that it is coprime to φ(n)?). I was wondering whether there is a proof that shows that a public key exponent e that is coprime with the totient of the modulus N will result in a unique answer when the cipher text is decrypted. Please help.

Answer 1

First, if $e$ is the public key exponent, then the private key exponent $d$ is the (unique) multiplicative inverse modulo $\phi(n)$, or alternatively $ed \equiv 1 \pmod{\phi(n)}$. The reason we want $d$ to have this property is because by Euler's theorem, which says $a^{\phi(n)} \equiv 1 \pmod{n}$ for essentially all values of $a$, so $(a^e)^d\equiv a^{ed} \equiv a \pmod{n}$ ensures we can always decrypt $a^e$ to $a$.

Now, how do we know $d$ is the unique multiplicative inverse of $e$? Because if there were another value of $d'$ such that $ed' \equiv 1 \pmod{\phi(n)}$, then $0 \equiv 1-1 \equiv (ed-ed') \equiv (d-d')e \equiv (d-d') \pmod{\phi(n)}$, where the last congruence holds because $\gcd(e,\phi(n))=1$. Thus, $d$ is unique.

Finally, if $\gcd(e,\phi(n)) \ne 1$, then $e$ does not have a multiplicative inverse mod $\phi(n)$. This is because, assuming $e^{-1}$ exists, then there exists some integer $k$ such that $k\phi(n) + ee^{-1} = 1$, which implies $\gcd(e,\phi(n)) = 1$.

In practice, $e$ not having a multiplicative inverse means that the sequence $a^e, a^{2e}, a^{3e}, \ldots$ may never attain the value $a$ modulo $n$.

As fgrieu points out, there are other values of $d$ that can serve as private key exponent (besides the multiplicative inverse of $e$) but we can always choose $d$ to be $e^{-1}$ and we are guaranteed that this choice will work with high probability.